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n 1 A 1 v ¯ 1 = n 2 A 2 v ¯ 2 , size 12{n rSub { size 8{1} } A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {}

where n 1 size 12{n rSub { size 8{1} } } {} and n 2 size 12{n rSub { size 8{2} } } {} are the number of branches in each of the sections along the tube.

Calculating flow speed and vessel diameter: branching in the cardiovascular system

The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0 L/min. The aorta has a radius of 10 mm. (b) Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0 L/min, the speed of blood in the capillaries is about 0.33 mm/s. Given that the average diameter of a capillary is 8.0 μ m , calculate the number of capillaries in the blood circulatory system.

Strategy

We can use Q = A v ¯ size 12{Q=A {overline {v}} } {} to calculate the speed of flow in the aorta and then use the general form of the equation of continuity to calculate the number of capillaries as all of the other variables are known.

Solution for (a)

The flow rate is given by Q = A v ¯ size 12{Q=A {overline {v}} } {} or v ¯ = Q πr 2 size 12{ {overline {v}} = { {Q} over {πr rSup { size 8{2} } } } } {} for a cylindrical vessel.

Substituting the known values (converted to units of meters and seconds) gives

v ¯ = 5.0 L/min 10 3 m 3 /L 1 min/ 60 s π 0 . 010 m 2 = 0 . 27 m/s . size 12{ { bar {v}}= { { left (5 "." 0`"L/min" right ) left ("10" rSup { size 8{ - 3} } `m rSup { size 8{3} } "/L" right ) left (1`"min/""60"`s right )} over {π left (0 "." "010 m" right ) rSup { size 8{2} } } } =0 "." "27"`"m/s"} {}

Solution for (b)

Using n 1 A 1 v ¯ 1 = n 2 A 2 v ¯ 1 size 12{n rSub { size 8{1} } A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {} , assigning the subscript 1 to the aorta and 2 to the capillaries, and solving for n 2 size 12{n rSub { size 8{2} } } {} (the number of capillaries) gives n 2 = n 1 A 1 v ¯ 1 A 2 v ¯ 2 . Converting all quantities to units of meters and seconds and substituting into the equation above gives

n 2 = 1 π 10 × 10 3 m 2 0.27 m/s π 4.0 × 10 6 m 2 0.33 × 10 3 m/s = 5.0 × 10 9 capillaries . size 12{n rSub { size 8{2} } = { { left (1 right ) left (π right ) left ("10" times "10" rSup { size 8{ - 3} } " m" right ) rSup { size 8{2} } left (0 "." "27"" m/s" right )} over { left (π right ) left (4 "." 0 times "10" rSup { size 8{ - 6} } " m" right ) rSup { size 8{2} } left (0 "." "33" times "10" rSup { size 8{ - 3} } " m/s" right )} } =5 "." 0 times "10" rSup { size 8{9} } " capillaries"} {}

Discussion

Note that the speed of flow in the capillaries is considerably reduced relative to the speed in the aorta due to the significant increase in the total cross-sectional area at the capillaries. This low speed is to allow sufficient time for effective exchange to occur although it is equally important for the flow not to become stationary in order to avoid the possibility of clotting. Does this large number of capillaries in the body seem reasonable? In active muscle, one finds about 200 capillaries per mm 3 size 12{"mm" rSup { size 8{3} } } {} , or about 200 × 10 6 size 12{"200" times "10" rSup { size 8{6} } } {} per 1 kg of muscle. For 20 kg of muscle, this amounts to about 4 × 10 9 size 12{4 times "10" rSup { size 8{9} } } {} capillaries.

Section summary

  • Flow rate Q size 12{Q} {} is defined to be the volume V size 12{V} {} flowing past a point in time t size 12{t} {} , or Q = V t size 12{Q= { {V} over {t} } } {} where V size 12{V} {} is volume and t size 12{t} {} is time.
  • The SI unit of volume is m 3 size 12{m rSup { size 8{3} } } {} .
  • Another common unit is the liter (L), which is 10 3 m 3 size 12{"10" rSup { size 8{ - 3} } `m rSup { size 8{3} } } {} .
  • Flow rate and velocity are related by Q = A v ¯ size 12{Q=A {overline {v}} } {} where A size 12{A} {} is the cross-sectional area of the flow and v ¯ size 12{ {overline {v}} } {} is its average velocity.
  • For incompressible fluids, flow rate at various points is constant. That is,
    Q 1 = Q 2 A 1 v ¯ 1 = A 2 v ¯ 2 n 1 A 1 v ¯ 1 = n 2 A 2 v ¯ 2 . size 12{ left none matrix { Q rSub { size 8{1} } =Q rSub { size 8{2} } {} ##A rSub { size 8{1} } {overline {v}} rSub { size 8{1} } =A rSub { size 8{2} } {overline {v}} rSub { size 8{2} } {} ## n rSub { size 8{1} } A rSub { size 8{1} } {overline {v}} rSub { size 8{1} } =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v}} rSub { size 8{2} }} right rbrace "." } {}

Conceptual questions

What is the difference between flow rate and fluid velocity? How are they related?

Many figures in the text show streamlines. Explain why fluid velocity is greatest where streamlines are closest together. (Hint: Consider the relationship between fluid velocity and the cross-sectional area through which it flows.)

Identify some substances that are incompressible and some that are not.

Problems&Exercises

What is the average flow rate in cm 3 /s size 12{"cm" rSup { size 8{3} } "/s"} {} of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L?

2.78 cm 3 /s size 12{"cm" rSup { size 8{3} } "/s"} {}

Questions & Answers

Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
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Renato
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Adin Reply
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
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biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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nano basically means 10^(-9). nanometer is a unit to measure length.
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Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
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Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
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Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
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s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
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Do you know which machine is used to that process?
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for screen printed electrodes ?
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s. Reply
of graphene you mean?
Ebrahim
or in general
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in general
s.
Graphene has a hexagonal structure
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Source:  OpenStax, Physics 101. OpenStax CNX. Jan 07, 2013 Download for free at http://legacy.cnx.org/content/col11479/1.1
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